Question

canada’s highest waterfall is della falls on vancouver island. an observer standing at the same level as the base of the falls views the top of the falls at an angle of elevation of 58°. when the observer moves 31 m closer to the base of the falls the angle of elevation increases to 61°. find the height of della falls. answer to the nearest metre.

Explanation:

Step1: Define Variables

Let \( h \) be the height of the falls, \( x \) be the initial distance from the observer to the base. So we have two right triangles:

• First triangle: \( \tan(58^\circ)=\frac{h}{x} \) ⇒ \( x = \frac{h}{\tan(58^\circ)} \)
• Second triangle: \( \tan(61^\circ)=\frac{h}{x - 31} \) ⇒ \( x - 31=\frac{h}{\tan(61^\circ)} \) ⇒ \( x=\frac{h}{\tan(61^\circ)}+ 31 \)

Step2: Equate the two expressions for \( x \)

\( \frac{h}{\tan(58^\circ)}=\frac{h}{\tan(61^\circ)}+ 31 \)

We know that \( \tan(58^\circ)\approx1.6003 \), \( \tan(61^\circ)\approx1.8040 \)

Substitute these values:
\( \frac{h}{1.6003}=\frac{h}{1.8040}+ 31 \)

Multiply through by \( 1.6003\times1.8040 \) to clear denominators:
\( 1.8040h=1.6003h + 31\times1.6003\times1.8040 \)

Step3: Solve for \( h \)

Subtract \( 1.6003h \) from both sides:
\( 1.8040h-1.6003h=31\times1.6003\times1.8040 \)
\( 0.2037h=31\times2.887 \) (approx, since \( 1.6003\times1.8040\approx2.887 \))
\( 0.2037h\approx89.497 \)
\( h=\frac{89.497}{0.2037}\approx439 \)

Answer:

\( \boxed{439} \)